mathematical induction2.2

a)when n = 1,
4(1^2) = (1/4)(1)(1+1)[1^2 + 5(1) + 2] = 4
Let when n = k the statement is true :
4(1^2) + 5(2^2) + 6(3^2) + ... + (n+3)n^2 = (1/4)n(n+1)[(n^2)+5n+2]
when n = k + 1:
(1/4)n(n+1)[(n^2)+5n+2] + (n+4)(n+1)^2
= (n+1)[(1/4)n(n^2 + 5n + 2) + (n+4)(n+1)]
= (n+1){(1/4)[n(n^2 + 5n + 2) + 4(n+4)(n+1)]}
= (1/4)(n+1)(n^3 + 5n^2 + 2n + 4n^2 + 20n + 16)
= (1/4)(n+1)(n^3 + 9n^2 + 22n + 16)
= (1/4)(n+1)(n+2)(n^2 + 7n + 8)
= (1/4)(n+1)(n+2)(n^2 + 2n + 1 + 5n + 7)
= (1/4)(n+1)(n+2)[(n+1)^2 + 5(n+1) + 2]
Hence it is true for n = k+1.
By mathematical induction it is true for all + integers.


b)
(1^2)x2 + (2^2)x3 + (3^2)x4 +...+ (n^2)(n+1)
= (1^2)x4 + (2^2)x5 + (3^2)x6 +...+ (n^2)(n+3) -
2[(1^2) + (2^2) + (3^2) +...+ (n^2)]
= (1/4)n(n+1)[(n^2)+5n+2] - 2(1/6)n(n+1)(2n+1)
= (1/4)n(n+1)[(n^2)+5n+2] - (1/3)n(n+1)(2n+1)
= (1/12)n(n+1)[3(n^2+5n +2) - 4(2n+1)]
= (1/12)n(n+1)[3n^2 + 15n + 6 - 8n - 4)
= (1/12)n(n+1)(3n^2 + 7n + 2)
= (1/12)n(n+1)(n+2)(3n+1)






2009-10-07 23:51:22 補充：
In part a),many (n) should be (k), sorry!